Zeros of the $n$-th derivative Suppose $f \in C^{\infty}(-1, 1)$ and $\sup_{(-1, 1)}|f(x)| \leq 1$. I need to prove that for every $n \in \mathbb{N}$ there is some $\alpha_n \in \mathbb{R}$ such that if $|f'(0)| \geq \alpha_n$ then there is at least $n-1$ solutions to $f^{(n)}(x) = 0$ on $(-1, 1)$.
Now this obviously needs some kind of induction. It is easy to prove this for $n=2$ (just pick some $A > 1$ and suppose $|f'(0)| \geq A$. Then if $|f'(x)|$ is $\geq A$ on all the $(-1, 1)$, then $|f(-1 + \epsilon) - f(1 - \epsilon)| > 2$ for some $\epsilon$ by the mean value theorem, which contradicts the supremum assumption, so there is points left and right to 0, where $|f'(0)|$ is less than $A$, which gives at least two points of different signs for $f''(x)$, and that, since $f''(x)$ is continuous, gives a zero somewhere inbetween).
Now if by induction we have $n-2$ zeros for $f^{n-1}(x)$, it is easy to prove the existence of at least $n-3$ zeros for $f^{n}(x)$ by the same mean value theorem. But how do I find 2 more by simply increasing $|f'(0)|$?
 A: This is not a solution, just an idea that I thought could lead to a complete proof, but have so far been unable to complete.
First, I'll reformulate the problem into a form I think is more easily addressed.
If $s=f'(0)$, let $g(x) = f(x/s)$. Then $g'(0)=1$. The statement that for some $\alpha>0$, if $|f'(0)|>\alpha$, then $f^{(n)}(x)$ has at least $n-1$ zeroes on $x\in[-1,1]$ can then be translated into the following statement in terms of $g(x)$:
Given $n$, there is some $\alpha>0$, so that for any $C^\infty$ function $g:[-\alpha,\alpha]\rightarrow[-1,1]$ with $g'(0)=1$, $g^{(n)}$ has at least $n-1$ zeroes.
The question is if, for a given $n$, is there some fixed $\alpha>0$, so the statement holds for all functions. However, if we don't require that we be able to fix a finite $\alpha>0$, we would get a somewhat simpler question:
If $g:\mathbb{R}\rightarrow[-1,1]$ is $C^\infty$, then $g^{(n)}$ has at least $n-1$ zeroes.
This, I think I should be able to prove using the argument below. Thus, the difficult part seems to be to get a fix on $\alpha$.
The idea of a proof of the simple statement could be as follows, where for convenience I'll just assume the zeroes are distinct:
Assuming $f^{(n)}$ has $n-1$ zeroes (or more), say at $x_1<\cdots<x_{n-1}$, the sign of $f^{(n+1)}(x_i)$ at these $n-1$ points will alternate, so $f^{(n+1)}$ must have zeroes at points $y_i$ inbetween these: ie $x_1<y_1<x_2<\cdots<y_{n-2}<x_{n-1}$. This gives us $n-2$ zeroes: we need two more.
Since $y_{n-2}<x_{n-1}$, if $f^{(n+1)}(x_{n-1})=0$, we have another zero. If not, it's non-zero: let's assume it's positive (same proof if negative). Also, to keep it simple, let's assume $x_{n-1}$ is the largest zero of $f^{(n)}$ and $y_{n-2}$ is the largest zero of $f^{(n+1)}$: if this is not the case, there will already be extra zeroes available.
So now we have $f^{(n)}(x_{n-1})=0$ and $f^{(n+1)}(y)>0$ (or $<0$ if we had assumed negative) for all $y>y_{n-1}$. This makes $f^{(n+1)}(x)>0$ and strictly increasing for $x>x_{n-1}$. In particular, if $f^{(n+1)}(x_{n+1}+\epsilon)=\delta$, this makes $f^{(n+1)}(x)\ge\delta$ for $x\ge x_{n-1}+\epsilon$, which makes $f(x)$ unbounded as $x\rightarrow\infty$. So, to avoid this, $f^{(n+1)}$ must have an additional zero at some $y_{n-1}>x_{n-1}$.
Same argument applies at the lower end, providing a zero at $y_0<x_1$.
I know there are a number of gaps in this proof to be filled in, and it could probably be formulated much more clearly, but this is the core idea.
